If B=.5(15-sqrt(197)), then A and B both satisfy x^2-15x+7=0. Let a(1),a(2),a(3), ... be an infinite sequence defined by a(1)=1, a(2)=15, and a(n)=15a(n-1)-7a(n-2). Now a(n)=a(n-1) mod 7 --> a(n)=1 mod 7.
If x^2-15x+7=0 then x^n=a(n)x-7a(n-1) for integer values of n>1. Proof (by induction): n=2 --> x^2=15x-7. Assume x^n=a(n)x-7a(n-1) --> x^(n+1)=a(n)x^2-7a(n-1)x = a(n)(15x-7)-7a(n-1)x = x(15a(n)-7a(n-1))-7a(n) = a(n+1)x-7a(n).
So A^n+B^n=a(n)A-7a(n-1)+a(n)B-7a(n-1) = a(n)(A+B)-14a(n-1) = 15a(n)-14a(n-1) = a(n) mod 7 = 1 mod 7 --> there exists positive integer k such that A^n+B^n = 7k+1 --> A^n=7k+1-B^n. But 0 < B < 1 --> 0 < B^n < 1 --> [A^n]=7k for n>1. Since [A]=14, the proof is complete. |
